What is the pH of a #1.4 * 10^-2 M# NaOH solution?

1 Answer
Dec 28, 2015

Answer:

#"pH" = 12.15#

Explanation:

Even before doing any calculations, you can say that since you're dealing with a strong base, the pH of the solution must be higher than #7#.

The higher the concentration of the base, the higher the pH will be.

In your case, you're dealing with a solution of sodium hydroxide, #"NaOH"#, a strong base that dissociates completely in aqueous solution to form sodium cations, #"Na"^(+)#, and hydroxide anions, #"OH"^(-)#

#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#

http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html

Notice that the salt dissociates in a #1:1# mole ratio with the hydroxide anions, you you can say that

#["OH"^(-)] = ["NaOH"] = 1.4 * 10^(-2)"M"#

Now, the pH of the solution is determined by the concentration of hydronium ions, #"H"_3"O"^(+)#. For aqueous solutions, the concentration of hydronium ions is related to the concentration of hydroxide ions by the ion product constant of water, #K_W#

#K_W = ["OH"^(-)] * ["H"_3"O"^(+)]#

At room temperature, you have

#K_W = 10^(-14)#

This means that the concentration of hydronium ions can be determined by using

#["H"_3"O"^(+)] = K_W/(["OH"^(-)])#

Plug in your values to get

#["H"_3"O"^(+)] = 10^(-14)/(1.4 * 10^(-2)) = 7.14 * 10^(-13)"M"#

The pH of the solution is equal to

#color(blue)("pH" = - log(["H"_3"O"^(+)])#

In your case,

#"pH" = -log(7.14 * 10^(-13)) = color(green)(12.15)#

As predicted, the pH is not only higher than #7#, but it is significantly higher than #7#.

Alternatively, you can use the pOH of the solution to find its pH. As you know,

#color(blue)("pOH" = - log(["OH"^(-)]))#

In your case,

#"pOH" = - log(1.4 * 10^(-2)) = 1.85#

You know that

#color(blue)("pH" + "pOH" = 14)#

and so, once again, you have

#"pH" = 14 - 1.85 = color(green)(12.15)#